College Algebra, 2013a

4.2 Graphs of Rational Functions 327
ˆ The behavior of y = g(x) as x →−∞: If imagine substituting x = −1 billion into
x−7
, we estimate
≈ very small (−). 12 Hence,
x−7
x 2 −x−6
x 2 −x−6
≈
−1 billion
1billion 2
g(x) =2− x − 7
x 2 ≈ 2 − very small (−) =2+verysmall(+)
− x − 6
In other words, as x →−∞, the graph of y = g(x) is a little bit above the line y =2.
x−7
ˆ The behavior of y = g(x) as x → ∞. To consider as x → ∞, we imagine
x 2 −x−6
substituting x = 1 billion and, going through the usual mental routine, find
x − 7
x 2 ≈ very small (+)
− x − 6
Hence, g(x) ≈ 2 − very small (+), in other words, the graph of y = g(x) is just below
the line y =2asx →∞.
On y = g(x), we have (again, without labels on the x-axis)
y
1
−1
x
6. Finally we construct our sign diagram. We place an ‘”’ abovex = −2 andx =3,anda‘0’
above x = 5 2
and x = −1. Choosing test values in the test intervals gives us f(x) is(+)on
the intervals (−∞, −2), ( −1, 5 (
2)
and (3, ∞), and (−) ontheintervals(−2, −1) and
5
2 , 3) .As
we piece together all of the information, we note that the graph must cross the horizontal
asymptote at some point after x = 3 in order for it to approach y = 2 from underneath. This
is the subtlety that we would have missed had we skipped the long division and subsequent
end behavior analysis. We can, in fact, find exactly when the graph crosses y = 2. As a result
of the long division, we have g(x) =2−
x−7
x−7
. For g(x) = 2, we would need
x 2 −x−6 x 2 −x−6 =0.
This gives x − 7 = 0, or x = 7. Note that x − 7 is the remainder when 2x 2 − 3x − 5 is divided
by x 2 − x − 6, so it makes sense that for g(x) to equal the quotient 2, the remainder from
the division must be 0. Sure enough, we find g(7) = 2. Moreover, it stands to reason that g
must attain a relative minimum at some point past x = 7. Calculus verifies that at x = 13,
we have such a minimum at exactly (13, 1.96). The reader is challenged to find calculator
windows which show the graph crossing its horizontal asymptote on one window, and the
relative minimum in the other.
12 In the denominator, we would have (1billion) 2 − 1billion − 6. It’s easy to see why the 6 is insignificant, but to
ignore the 1 billion seems criminal. However, compared to (1 billion) 2 , it’s on the insignificant side; it’s 10 18 versus
10 9 . We are once again using the fact that for polynomials, end behavior is determined by the leading term, so in
the denominator, the x 2 term wins out over the x term.